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Chapter 24: Problem 75

Veronique is nearsighted; she cannot sce clearly anything more than $6.00 \mathrm{m}$ away without her contacts. One day she doesn't wear her contacts; rather, she wears an old pair of glasses prescribed when she could see clearly up to \(8.00 \mathrm{m}\) away. Assume the glasses are \(2.0 \mathrm{cm}\) from her eyes. What is the greatest distance an object can be placed so that she can see it clearly with these glasses?

Short Answer

Expert verified
Answer: Approximately 10.43 meters.

Step by step solution

01

Find the focal length of the old glasses

We know that with the old glasses, Veronique could see clearly up to 8 meters away (\(d_i = 8~m\)). We also know that without the glasses her far point is 6 meters away (\(d_o = 6~m\)). Using the lens equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) \(\frac{1}{f} = \frac{1}{6} + \frac{1}{8}\) \(f = \frac{1}{\frac{1}{6} + \frac{1}{8}} = 24/5~m = 4.8~m\)
02

Calculate the new image distance with the glasses on the face

Since the glasses are 2 cm from her eyes, the new image distance \(d_i'\) will be 2 cm less than the original image distance: \(d_i' = d_i - 0.02 = 8 - 0.02 = 7.98~m\)
03

Calculate the new object distance using the lens equation

Now that we know the focal length and the new image distance, we can find the new object distance \(d_o'\) using the lens equation: \(\frac{1}{f} = \frac{1}{d_o'} + \frac{1}{d_i'}\) \(\frac{1}{4.8} = \frac{1}{d_o'} + \frac{1}{7.98}\) Now, solve for \(d_o'\): \(d_o' = \frac{1}{\frac{1}{4.8} - \frac{1}{7.98}} \approx 10.43 ~m\)
04

Conclusion

The greatest distance an object can be placed so that Veronique can see it clearly with the old glasses is approximately 10.43 meters away.

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Most popular questions from this chapter

A microscope has an eyepicce that gives an angular magnification of 5.00 for a final image at infinity and an objective lens of focal length $15.0 \mathrm{mm}\(. The tube length of the microscope is \)16.0 \mathrm{cm} .$ (a) What is the transverse magnification due to the objective lens alone? (b) What is the angular magnification due to the microscope? (c) How far from the objective should the object be placed?
You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length $3.70 \mathrm{cm}\( located \)6.00 \mathrm{cm}$ to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high, what is the image height?
Kim says that she was less than 10 ft away from the president when she took a picture of him with her \(50-\mathrm{mm}\) focal length camera lens. The picture shows the upper half of the president's body (or \(3.0 \mathrm{ft}\) of his total height). On the negative of the film, this part of his body is $18 \mathrm{mm}$ high. How close was Kim to the president when she took the picture?
A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.
Two converging lenses, separated by a distance of \(50.0 \mathrm{cm},\) are used in combination. The first lens, located to the left, has a focal length of \(15.0 \mathrm{cm} .\) The second lens, located to the right, has a focal length of \(12.0 \mathrm{cm} .\) An object, \(3.00 \mathrm{cm}\) high, is placed at a distance of \(20.0 \mathrm{cm}\) in front of the first lens. (a) Find the intermediate and final image distances relative to the corresponding lenses. (b) What is the total magnification? (c) What is the height of the final image?
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